The distance between the foci of the hyperbola $x^2-3 y^2-4 x-6 y-11=0$ is
The distance between the foci of the hyperbola $x^2-3 y^2-4 x-6 y-11=0$ is
- $4$
- $6$
- $8$
- $10$
Solution
Given, equation of hyperbola is
$
\begin{gathered}
x^2-3 y^2-4 x-6 y-11=0 \\
\Rightarrow \quad\left(x^2-4 x+4\right)-3\left(y^2+2 y+1\right)-11 \\
=4-3 \\
\Rightarrow \quad(x-2)^2-3(y+1)^2=12 \\
\Rightarrow \quad \frac{(x-2)^2}{12}-\frac{(y+1)^2}{4}=1 \\
\text { Now, } \quad e=\sqrt{1+\frac{4}{12}}=\frac{2}{\sqrt{3}}
\end{gathered}
$
Now,
$
e=\sqrt{1+\frac{4}{12}}=\frac{2}{\sqrt{3}}
$
$\therefore$ Distance between foci
$
=2 a e=2 \times \sqrt{12} \times \frac{2}{\sqrt{3}}=8
$
Asked in: AP EAMCET 2008
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